y^3 + xy^2 + x^2y + 64 = 0
两边求导:
3y^2y' + y^2 + 2xyy' + 2xy + x^2y' = 0
∴y' = -(y^2 + 2xy)/(3y^2 + 2xy + x^2)
y(0) = 4
y'(0) = -4^2/3*4^2 = -1/3
y" = [-(y^2 + 2xy)'(3y^2 + 2xy + x^2)+(y^2 + 2xy)(3y^2 + 2xy + x^2)']/(3y^2 + 2xy + x^2)^2
= [-(2yy'+2y+2xy')(3y^2 + 2xy + x^2)+(y^2 + 2xy)(6yy'+2y+2xy'+2x)]/(3y^2 + 2xy + x^2)^2
= [4xy^2y'+2x^2yy'-2x^3y'-4y^3-2xy^2+2x^2y]/(3y^2 + 2xy + x^2)^2
= [-6x^2y^3+6x^4y-12y^5-18xy^4-4xy^3+2x^2y^2]/(3y^2 + 2xy + x^2)^3
y“(0) = 0
答题不易,请及时采纳,谢谢!